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The derivative of a function at any point is the limit of the ratio of the change of a function for an infinitely small variation of the input when the variation of the input tends to 0. This concept and the graphical interpretation of a derivative helps us to frame the definition of derivatives.

Derivatives are the central topic of differential calculus. The definition of a derivative is defined as a limit of the differential quotient.

**Definition of Derivative**

The derivative of ‘f’ at x is given by f'(x) = $\lim_{h \to 0}$ $\frac{f(x+h)-f(x)}{h}$

The derivative represents the instantaneous rate of change of the function. Thus geometrically f’(x) represents the slope of the tangent at x. The derivative of the position function is the velocity function and in turn, the derivative of the velocity function is the acceleration function.

Derivatives are the central topic of differential calculus. The definition of a derivative is defined as a limit of the differential quotient.

The derivative of ‘f’ at x is given by f'(x) = $\lim_{h \to 0}$ $\frac{f(x+h)-f(x)}{h}$

The derivative represents the instantaneous rate of change of the function. Thus geometrically f’(x) represents the slope of the tangent at x. The derivative of the position function is the velocity function and in turn, the derivative of the velocity function is the acceleration function.

Let us learn what are derivatives:

Let f(x) be function of x. Consider a small increase h in the value of x.

The value of the function with this increase is f(x + h)

The change in value of the function is f(x + h) – f(x)

The rate of change of the function at this interval is [f(x + h) – f(x)]/h

**Let us study the above situation graphically,**

Let the curve represents a function f(x). Consider two points A and B at x = x and at x = x + h. The corresponding values of the function are f(x) and f(x + h). Let *t *and *t’* are the tangents at A and B.

As *h* tends to 0, the point A will merge with the point B and the tangent *t* will merge with the tangent *t’*.

*Graphically, the derivative of a function at any point is the slope of the tangent to the graph of the function at that point.*

Therefore the derivative at point B is nothing but the limit of [f(x + h) - f(x)]/h, as 'h' tends to 0.

This is the fundamental definition of a derivative.

For a real valued function f which has a point a in its domain of definition derivative of f at a is given by

$\mathop{\lim}\limits_{h \to \ 0}$ $\frac{f(a+h)-f(a)}{h}$

For a real valued function f which has a point a in its domain of definition derivative of f at a is given by

$\mathop\lim\limits_{h \to \ 0}$ $\frac{f(a+h)-f(a)}{h}$

Provided this limit exists. Derivative of f(x) at a is denoted by f’(a).

Geometrical representation of derivative of a function at a point is given below –

Let y = f(x) be a function and let M = (a, f (a)) and N = (a + h, f (a +h) ) be two point close to each other on graph of this function.

As we have written above –

f’(a) = $\mathop\lim\limits_{h \to \ 0}$ $\frac{f(a+h)-f(a)}{h}$

From the triangle LMN, it is clear that ratio whose limit we are taking is precisely equal to tan(NML) which is the slope of the chord MN. In the limiting process, as h tends to 0, the points N tends to M and we have –Let y = f(x) be a function and let M = (a, f (a)) and N = (a + h, f (a +h) ) be two point close to each other on graph of this function.

As we have written above –

$\mathop\lim\limits_{h \to \ 0}$ $\frac{f(a+h)-f(a)}{h}$ = $\mathop\lim\limits_{N \to \ M}$ $\frac{QR}{PR}$

Thus the limit turns out equal to be equal to slope of the tangent. Hence

f’(a) = tan θ

The formulas for derivatives of common functions are found using the limit definition of the derivative.

For example let us derive the derivative of x using the first principle,

f(x) = x

f’(x) = $\lim_{h\rightarrow 0}\frac{x+h-x}{h}$ By the Limit definition of the derivative

= $\lim_{h\rightarrow 0}\frac{x+h-x}{h}$ = $\lim_{h\rightarrow 0}\frac{h}{h}$ = $\lim_{h\rightarrow 0}1$ = 1

The constant function has the same function values for all values of x. Hence the derivative of a constant function = 0. So we can say the derivative of 1 = 0.

**A list of derivatives of common functions is given in the table below:**

For example let us derive the derivative of x using the first principle,

f(x) = x

f’(x) = $\lim_{h\rightarrow 0}\frac{x+h-x}{h}$ By the Limit definition of the derivative

= $\lim_{h\rightarrow 0}\frac{x+h-x}{h}$ = $\lim_{h\rightarrow 0}\frac{h}{h}$ = $\lim_{h\rightarrow 0}1$ = 1

The constant function has the same function values for all values of x. Hence the derivative of a constant function = 0. So we can say the derivative of 1 = 0.

Table for Derivative formulas | |

Function - f(x) | Derivative function - f’(x) |

x_{n} |
nx^{n-1} |

Any constant | 0 |

Derivative of exponential function ex | ex |

Derivative of exponential function a^{x} |
ln(a)ax |

Derivative of logarithmic function ln(x) | $\frac{1}{x}$ |

Derivatives of Trigonometric functions -f(x) | Derivative of the function f'(x) |

Sin x | cos x |

Cos x | -Sin x |

Tan x | sec2x |

Csc x | -cot x csc x |

Sec x | Sec x tan x |

Cot x | -csc2x |

Derivatives in Inverse trigonometric functions f(x) | Derivative function -f'(x) |

Sin^{-1}x |
$\frac{1}{\sqrt{1-x^2}}$ |

Cos^{-1}x |
$\frac{-1}{\sqrt{1-x^2}}$ |

Tan^{-1}x |
$\frac{1}{1+x^2}$ |

Csc^{-1}x |
$\frac{-1}{x\sqrt{x^2-1}}$ |

Sec^{-1}x |
$\frac{1}{x\sqrt{x^2-1}}$ |

Cot^{-1}x |
$\frac{-1}{1+x^2}$ |

- Product Rule
- Quotient Rule
- Chain Rule

The derivative examples are shown along with each rule applied.

The product rule is used when the function is given as the product of two functions

**Derivative Product Rule:**

**An example for finding the derivative using the product rule:**

Find the derivative of f(x) = x^{2} sin x

f(x) is a product of two functions , where u(x) = x^{2} and v(x) = sin x

u’(x) = 2x and v’(x) = cos x.

f’(x) = udv + vdu = x^{2} cos x + 2x sin x. (Using the derivative formulas given in the table)

The quotient rule is applied when the function is given as a quotient of two functions

**Quotient Rule Derivative:**

If u and v are differentiable at x, and if v(x) 0, then the quotient u/v is differentiable at x and

$\frac{du}{dx}(\frac{u}{v})$ = $\frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$

**Derivative Example using Quotient rule**

Find the derivative of $\frac{e^x}{x^2+2}$

u(x) = e^{(x)} and v(x) = x^{2} +2 u’(x) = e^{x} and v’(x) = 2x

Hence the derivative =$\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}$ = $\frac{(x^2+2)e^x-e^x(2x)}{(x^2+2)^2}$ = $\frac{e↑x(x↑2-2x+2)}{\square (x↑2+2)\square ↑2}$

The chain rule is used in function compositions. That is when one function contains another function. The chain rule is generally used in a sequence of chains.

**Chain Rule**

The working method of the chain rule is easy. The derivative of the outer function is multiplied by the derivative of the inner function taking the variables in the order.

**Derivative example using the chain rule:**

Find the derivative of $e^{tanx}$

The outer function is the exponential function of the form e^{x} and the inner function is tan x

f’(x) = derivative of the outer x derivative of the inner

= $e^{tanx}.sec^2x$

The product rule is used when the function is given as the product of two functions

If ‘u’ and ‘v’ are differentiable at x, then

$\frac{d}{dx(uv)}$ = $\frac{dv}{u.dx}+\frac{du}{v.dx}$

Find the derivative of f(x) = x

f(x) is a product of two functions , where u(x) = x

u’(x) = 2x and v’(x) = cos x.

f’(x) = udv + vdu = x

The quotient rule is applied when the function is given as a quotient of two functions

If u and v are differentiable at x, and if v(x) 0, then the quotient u/v is differentiable at x and

$\frac{du}{dx}(\frac{u}{v})$ = $\frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$

Find the derivative of $\frac{e^x}{x^2+2}$

u(x) = e

Hence the derivative =$\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}$ = $\frac{(x^2+2)e^x-e^x(2x)}{(x^2+2)^2}$ = $\frac{e↑x(x↑2-2x+2)}{\square (x↑2+2)\square ↑2}$

The chain rule is used in function compositions. That is when one function contains another function. The chain rule is generally used in a sequence of chains.

If F(x) = f(g(x)) and if f(x) and g(x) are differentiable, then

F’(x) = f’(g(x))g’(x)

The working method of the chain rule is easy. The derivative of the outer function is multiplied by the derivative of the inner function taking the variables in the order.

Find the derivative of $e^{tanx}$

The outer function is the exponential function of the form e

f’(x) = derivative of the outer x derivative of the inner

= $e^{tanx}.sec^2x$

The derivative f’(x) is the limit of the difference quotient $\frac{f(x+h)-f(x)}{h}$. If a and b are the end points of an interval, where the function is continuous, the difference quotient $\frac{f(b)-f(a)}{b-a}$ gives the slope of the secant line joining a and b. The mean value theorem states that if the function is differentiable in (a, b), then there is a point c$\varepsilon $ (a, b), where the tangent is parallel to the line joining a and b. To put it in mathematical terms f'(c) = $\frac{f(b)-f(a)}{b-a}$

- Find the first derivative f’(x)
- Choose two points a, b around the critical point x = c such that a < c < b
- Find the signs of f’(a) and f’(b).
- If f’(a) is positive and f’(b) is negative, the function is increasing for values of x<c and decreasing for values of x > c. Hence x = c is a local maximum.
- If f’(a) is negative and f’(b) is positive, the function is decreasing for values of x<c and increasing for values of x > c. Hence x = c is a local minimum.

If the derivatives are differentiable, they can be differentiated successively to get higher order derivatives.

The second derivative f”(x) is got by differentiating f’(x). Geometrically if the first derivative represents the slope of the tangent at x, the second derivative gives the concavity of the function. Similarly, if the first derivative of the position function gives the velocity, the second derivative got by differentiating the velocity function gives the acceleration. When differentiated further, the derivative of acceleration function tells us about the behavior of acceleration.

The notation for second derivative is $\frac{d^2y}{d^2x}$ or f”(x)

The second derivative f”(x) is got by differentiating f’(x). Geometrically if the first derivative represents the slope of the tangent at x, the second derivative gives the concavity of the function. Similarly, if the first derivative of the position function gives the velocity, the second derivative got by differentiating the velocity function gives the acceleration. When differentiated further, the derivative of acceleration function tells us about the behavior of acceleration.

The notation for second derivative is $\frac{d^2y}{d^2x}$ or f”(x)

If cos x is the derivative of sin x, then sin x is obtained by the reversal of differentiation on cos x. Sin x is the result of anti differentiation on cos x

**Antiderivative – Definition**

A function of the form sin x +c will yield cos x on differentiation, as the derivative of the constant is zero. Hence the constant is added to give the general antiderivative.

A function F(x) is called an antiderivative of f(x) on an interval if F’(x) = f(x) for all x in the interval.

The most general antiderivative of f(x) is F(x) + C where ‘c’ is an arbitrary constant.

A function of the form sin x +c will yield cos x on differentiation, as the derivative of the constant is zero. Hence the constant is added to give the general antiderivative.

In advanced multi-variable calculus, we come across functions with more than one variable. Partial derivatives are found by differentiating the function with respect to one of the variables while freezing the other variables as constants.

For example: if z = 2x + 3y^{2}, the partial derivative of Z with respect to x and y represented by the notations $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$. They are read as ‘doh z over doh x’ and ‘doh z over doh y’. The computation of partial derivatives is shown below.

$\frac{\partial z}{\partial x}$=$\frac{\partial }{\partial x}$(2x ) + (3y^{2}) = 2 + 0 = 2

$\frac{\partial z}{\partial y}$= $\frac{\partial }{\partial y}$(2x) + (3y^{2}) = 0 + 3(2y) = 6y.

For example: if z = 2x + 3y

$\frac{\partial z}{\partial x}$=$\frac{\partial }{\partial x}$(2x ) + (3y

$\frac{\partial z}{\partial y}$= $\frac{\partial }{\partial y}$(2x) + (3y

In single variable calculus the rate of change of the function is calculated along the direction of the tangent at the point. But in multivariable calculus, the change can take place in many directions. Hence while specifying the rate of change, it is necessary to state the direction in which the change takes place also. The rate of change in a given direction is known as the directional derivative. The formula for directional derivative in the direction of a unit vector is defined using the limit definition as follows:

The directional derivative of ‘f’ at (x

D

Derivative RulesImplicit DifferentiationDerivative of Trigonometric Functions Derivatives of Exponential FunctionsPartial DerivativeDirectional Derivative Derivative Chain RuleInstantaneous Rate of ChangeMean Value Theorem Derivatives of Inverse Trigonometric FunctionsDerivative of LogSecond Derivative AntiderivativesDerivative Table